2X of 0.3 + 2 = 1.4-3x of 0.2

2X of 0.3 + 2 = 1.4-3x of 0.2

2x+2/3=1.4-3x/2
4x+4=4.2-9x
13x=0.2
x=2/130
2.2x-0.5-0.3x
It's better to have a process
3x-0.6 in 0.2 = 2x-1.5-0.1 in 0.5 + 4.2
3x/(0.2)-0.6=2x/(0.5)-1.5-x/(0.1)+4.2
3x/(0.2)-2x/(0.5)+x/(0.1)=-1.5+4.2+0.6
15x-4x+10x=4.2+0.6-1.5
21x=3.3
x=11/70
2X in 0.3 minus 1.6 in 0.6 minus 3x is equal to 31x in 3 plus 8
A:
2X in 0.3 minus 1.6 in 0.6 minus 3x is equal to 31x in 3 plus 8
2x/0.3-(1.6-3x)/0.6=(31x+8)/3
(4x-1.6+3x)/0.6=(31x+8)/3
5(7x-1.6)=31x+8
35x-8=31x+8
35x-31x=8+8
4x=16
X=4
20x/3-(8-15x)/3=(31x+8)/3
20x-8+15x=31x+8
4x=16
X=4
1.4x+3=3x+5 2.12-4x=3x-2
1.4x+3=3x+5 2.12-4x=3x-2 3.3(2x-1)=4(3-x) 4.12-2(3x-4)=x 5.13-2(2x-3)=5-(x-2)
        4x+3=3x+5   4x-3x=5-3x=2               2.    &n...
X=2X=2X=1.5X=20/7X=4
As shown in the figure, the image of the first-order function passing through the point Q (0,3.5) intersects the image of the positive scale function y = 2x at the point P. the equation that can represent the image of the first-order function is ()
A. 3x-2y+3.5=0B. 3x-2y-3.5=0C. 3x-2y+7=0D. 3x+2y-7=0
Let y = KX + B. ∵ the line passes through points P (1,2) and Q (0,3.5), K + B = 2B = 3.5, and the solution is k = − 1.5b = 3.5. Therefore, the analytic expression of the linear function is y = - 1.5x + 3.5, that is, 3x + 2y-7 = 0
Given the set a = {x | log2 (x) ≤ 2}, B = (negative infinity, a), if a contains and B, then the value range of real number a is (C, positive infinity), find the value of C
log2(x)≤2=log2(4)
x≤4
A is contained in B
Then X4
So C = 4
A is equivalent to (0,4) a is contained in B, which means that a is greater than 4, C is 4
Solving inequality 3x - / X-5/
discuss
x> 5:3x - (X-5)
When x > = 5
2X+5